How does the hardy weinberg formula work?

Question

Not so much explaining as actual problem solving. For example,

If someone were to tell you that there was a population of 100 people, 98 of them having attached earlobes (rr). What would be the:
P:
Q:
f(Homozygous Dominant):
f(Homozygous Recessive):
#(Homozygous Dominant):
#(Homozygous Recessive):

Thank you, a definite 10 points for a valid answer.

Answer 1

This link will explain how to set up this type of problem: http://www.biologycorner.com/bio4/notes/hardy_weinberg.php

You were given that the number of homozygous recessive individuals was 98 (out of a sample of 100). So, 98/100 = 0.98 (this is your q2). So 2/100 would be homozygous dominant.

The square root of that (q) = 0.9899

1 - 0.9899 = 0.0100 0.0100 = p

p squared = 0.0001 This means 1/100th of a person would homozygous dominant (better to multiply through by the 100, and say 1 in 10,000 are this).

although you didn't ask, 2pq = % is the number that are heterozygous, so 2*0.9899*0.01 = 0.0197

So, if you had 100 randomly selected people, 98 should be homozygous recessive, almost 2 should be heterozygous, and a very slim change that any would be homozygous dominant

p = % of population with the dominant allele - since 98 have the recessive trait, p for your problem would be 2% (0.02) - this would be the frequency of the dominant allele

q = % of population with the recessive allele - since 98 have the recessive trait, q for your problem would be 98% (0.98) - this would be the frequency of the recessive allele